Everything here is based on the binomial expansion . This can be used in two ways to expand , where a is a constant.

1. (expansion in positive powers of z);

2. (expansion in negative powers of z).

The expansion in positive powers of z converges when |z|<|a|. The expansion in negative powers of z converges when |z|>|a|.

To get the Laurent expansion for f(z) = 1/((z+1)(z+2)) that converges for z=7/2, use partial fractions and then, for each of the partial fraction terms, use whichever of the above series converges for z=7/2.

Edit. After reading the question more carefully (something it always pays to do ), I see that you want a series that converges in an annulus. So you want the singularities at z=-1 and z=-2 to be on the inner and outer circles of the annulus. You also want the point z=7/2 to be inside the annulus. So (as indicated in the hint) it would be a good idea to centre the annulus at the point z=1. Then the inner radius will be 2 (the distance from z=1 to z=-1), the outer radius will be 3 (the distance from z=1 to z=-2) and the distance from z=1 to z=7/2 is 5/2 which lies between 2 and 3.

The way to do the question is then to make a substitution w=z-1, so that the function becomes . Then use the procedure that I described above to get the Laurent series for this function of w that converges when w=5/2. Finally, substitute back again to get a series in positive and negative powers of (z-1).

Last edited by Opalg; November 10th 2008 at 04:04 AM.
Reason: I read the question again!